Finding the Area of a Sector using Trigonometric Functions

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The discussion focuses on calculating the area of a sector using trigonometric functions, specifically addressing part (c) of a problem. One participant shares their solution, arriving at an area of approximately 80.84 cm² using the sine function. Another participant suggests an alternative approach using the cosine function, noting its relation to the sine of 1 radian. A minor correction is made regarding the area calculation, adjusting it to 80.83 cm². The conversation highlights different methods for solving the same problem in trigonometry.
chwala
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Homework Statement
See attached
Relevant Equations
Area of sector
My interest is on part (c) only.

1648079111365.png


Wow, this was a nice one! boggled me a little bit anyways; my last steps to solution,
##A= \dfrac{1}{2} ×16.8319^2 × 0.5707=80.84cm^2## bingo!
Any other approach apart from using sine?
Cheers guys
 
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chwala said:
Any other approach apart from using sine?

Yes ! Using the cosine of angle FEH (which happens to be the sine of 1 radian, but that may well be a coincidence :smile: )

Your calculator needs new batteries: not 80.84 but 80.83 :wink: !

##\ ##
 

Thread 'Find the polar form of a complex number'

The working out suggests first equating ## \sqrt{i} = x + iy ## and suggests that squaring and equating real and imaginary parts of both sides results in ## \sqrt{i} = \pm (1+i)/ \sqrt{2} ## Squaring both sides results in: $$ i = (x + iy)^2 $$ $$ i = x^2 + 2ixy -y^2 $$ equating real parts gives $$ x^2 - y^2 = 0 $$ $$ (x+y)(x-y) = 0 $$ $$ x = \pm y $$ equating imaginary parts gives: $$ i = 2ixy $$ $$ 2xy = 1 $$ I'm not really sure how to proceed from here.
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